在where子句中的Oracle日期比较 [英] Oracle date comparison in where clause

问题描述

例如，我有一个学生表，带有一个DOJ(加入日期)列，其类型现在设置为DATE，因为我以dd-mon-yy格式存储了记录.

For eg I have a student table with a DOJ(date of joining) column with its type set as DATE now in that I have stored records in dd-mon-yy format.

我在运行时有一个IN参数，日期以字符串形式传递，其格式为dd/mm/yyyy.如何比较和获取日期结果?

I have an IN param at runtime with date passed as string and its in dd/mm/yyyy format. How do I compare and fetch results on date?

我想获取每个数据库表学生的DOJ为25-AUG-92的学生的记录数，但是我以dd/mm/yyyy格式在IN参数中获取varchar的日期，请指导.

I want to fetch count of records of students who have DOJ of 25-AUG-92 per my database table student, but I am getting date as varchar in dd/mm/yyyy format in an IN param, kindly please guide.

我尝试了多个选项，例如trunc，to_date，to_char，但不幸的是似乎没有任何作用.

I have tried multiple options such as trunc, to_date, to_char but, unfortunately nothing seems to work.

推荐答案

我现在有一个学生表，带有一个DOJ(加入日期)列，其类型设置为DATE，因为我已经以dd-mon-yy格式存储了记录.

I have a student table with a DOJ(date of joining) column with its type set as DATE now in that I have stored records in dd-mon-yy format.

不太完全，DATE数据类型没有格式；它以 7字节(年为2字节，月，日，小时，分钟和秒均为1字节).您正在使用的用户界面(即SQL/PLUS，SQL Developer，Toad等)将处理DATE的格式，从其二进制格式到人类可读的格式.在SQL/Plus(或SQL Developer)中，此格式基于 NLS_DATE_FORMAT会话参数.

Not quite, the DATE data-type does not have a format; it is stored internally in tables as 7-bytes (year is 2 bytes and month, day, hour, minute and second are 1-byte each). The user interface you are using (i.e. SQL/PLUS, SQL Developer, Toad, etc.) will handle the formatting of a DATE from its binary format to a human readable format. In SQL/Plus (or SQL Developer) this format is based on the NLS_DATE_FORMAT session parameter.

如果仅使用日，月和年输入DATE，则时间部分(可能)将设置为00:00:00(午夜).

If the DATE is input using only the day, month and year then the time component is (probably) going to be set to 00:00:00 (midnight).

我在运行时有一个IN参数，日期以字符串或varchar的形式传递，其格式为dd/mm/yyyy.如何比较和获取日期结果??

I have an IN param at runtime with date passed as string or say varchar and its in dd/mm/yyyy format. How do I compare and fetch results on date.?

假设您的DOJ列的时间部分始终是午夜，那么:

Assuming the time component for you DOJ column is always midnight then:

SELECT COUNT(*)
FROM   students
WHERE  doj = TO_DATE( your_param, 'dd/mm/yyyy' )

如果不总是午夜，那么:

If it isn't always midnight then:

SELECT COUNT(*)
FROM   students
WHERE  TRUNC( doj ) = TO_DATE( your_param, 'dd/mm/yyyy' )

或:

SELECT COUNT(*)
FROM   students
WHERE  doj >= TO_DATE( your_param, 'dd/mm/yyyy' )
AND    doj <  TO_DATE( your_param, 'dd/mm/yyyy' ) + INTERVAL '1' DAY

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